[R] regression coefficients

[Pfaff, Bernhard]

beside ?anova(), you might also be interested in considering the j-test, a
quick google retrieved:

http://support.sas.com/rnd/app/examples/ets/spec/

for an elucidation of the test problem.

HTH,
Bernhard

-----Original Message-----
From: Prof Brian Ripley [mailto:ripley at stats.ox.ac.uk]
Sent: 20 May 2003 17:51
To: Spencer Graves
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] regression coefficients


Why is s assumed known and common to the k groups?  I doubt if that is 
what was meant (although it was too imprecise to be at all sure).

If `common' is a viable assumption, you can just fit a model with by-group
regressions vs one with a common regression (which seems to be what you
are testing) and use anova().

If not, the case k=2 encompasses the Welch t-test so exact distribution
theory is not going to be possible, but by fitting a common model and
three separate models and summing the -2log-lik for the latter you can
easily get the LT test and refer it to its `standard' (asymptotic)
Chi-squared distribution.

On Tue, 20 May 2003, Spencer Graves wrote:

> 	  I don't know of a simply function to do what you want, but I can
give 
> you part of the standard log(likelihood ratio) theory:
> 
> 	  Suppose b[i]|s ~ N.r(b, s^2*W[i]), i = 1, ..., k.  Then the 
> log(likelihood) is a sum of k terms of the following form:
> 
> 	  l[i] = (-0.5)*(r*log(2*pi*s^2)+log|W[i]|
> 	      +(s^-2)*t(b[i]-b)%*%solve(W[i]%*%(b[i]-b)
> 
> By differentiating with respect to b and setting to 0, we get the 
> maximum likelihood estimate for b as follows:
> 
> 	  b.hat = solve(sum(solve(W[i]))%*%sum(solve(W[i])%*%b[i])
> 
> In words:  b.hat = weighted average with weights inversely proportional 
> to the variances.  Then log(likelihood ratio) is as follows:
> 
> 	  log.LR = sum((s^-2)*t(b[i]-b.hat)%*%solve(W[i])%*%(b[i]-b.hat))
> 
> This problem should be in most good books on multivariate analysis.  I 
> would guess that log.LR probably has an F distribution with numerator 
> degrees of freedom = r*(k-1) and with denominator degrees of freedom = 
> degrees of freedom in the estimate of s.  However, I don't remember for 
> sure.  It's vaguely possible that this is an "unsolved" problem.  In the 
> latter case, you should have all the pieces here to generate a Monte 
> Carlo.

You have assumed s is known, in which case it is a Chi-squared 
distribution.  If s is unknown, you need to maximize over it to get an LR 
test (separately under the null and the alternative).

> lamack lamack wrote:
> > dear all, How can I compare regression coefficients across three (or 
> > more) groups?

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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